A276084 a(n) = Number of trailing zeros in primorial base representation of n (A049345); largest k such that A002110(k) divides n.

0, 1, 0, 1, 0, 2, 0, 1, 0, 1, 0, 2, 0, 1, 0, 1, 0, 2, 0, 1, 0, 1, 0, 2, 0, 1, 0, 1, 0, 3, 0, 1, 0, 1, 0, 2, 0, 1, 0, 1, 0, 2, 0, 1, 0, 1, 0, 2, 0, 1, 0, 1, 0, 2, 0, 1, 0, 1, 0, 3, 0, 1, 0, 1, 0, 2, 0, 1, 0, 1, 0, 2, 0, 1, 0, 1, 0, 2, 0, 1, 0, 1, 0, 2, 0, 1, 0, 1, 0, 3, 0, 1, 0, 1, 0, 2, 0, 1, 0, 1, 0, 2, 0, 1, 0, 1, 0, 2, 0, 1, 0, 1, 0, 2, 0, 1, 0, 1, 0, 3

Offset: 1

Author

Antti Karttunen, Aug 22 2016

Keywords

• nonn
• base

Comments

Terms begin from a(1)=0 because for zero the count is ambiguous.

From Amiram Eldar, Mar 10 2021: (Start)

The asymptotic density of the occurrences of k is (prime(k+1)-1)/A002110(k+1).

The asymptotic mean of this sequence is Sum_{k>=1} 1/A002110(k) = 0.705230... (A064648). (End)

Examples

			For n=24, which is "400" in primorial base (as 24 = 4*(3*2*1) + 0*(2*1) + 0*1, see A049345), there are two trailing zeros, thus a(24) = 2.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 1..2310
• Index entries for sequences related to primorial base

Crossrefs

Cf. A000040, A001221, A002110, A049345, A053589, A064648, A340346.

One less than A257993.

Differs from the related A230403 for the first time at n=24.

Programs

Mathematica

Table[If[# == 0, 0, j = #; While[! Divisible[n, Times @@ Prime@ Range@ j], j--]; j] &@ If[OddQ@ n, 0, k = 1; While[Times @@ Prime@ Range[k + 1] <= n, k++]; k], {n, 120}] (* or *)
nn = 120; b = MixedRadix[Reverse@ Prime@ Range@ PrimePi[nn + 1]]; Table[Length@ TakeWhile[Reverse@ IntegerDigits[n, b], # == 0 &], {n, nn}] (* Version 10.2, or *)
f[n_] := Block[{a = {{0, n}}}, Do[AppendTo[a, {First@ #, Last@ #} &@ QuotientRemainder[a[[-1, -1]], Times @@ Prime@ Range[# - i]]], {i, 0, #}] &@ NestWhile[# + 1 &, 0, Times @@ Prime@ Range[# + 1] <= n &]; Rest[a][[All, 1]]]; Table[Length@ TakeWhile[Reverse@ f@ n, # == 0 &], {n, 120}] (* Michael De Vlieger, Aug 30 2016 *)

Python

from sympy import nextprime, primepi
def a053669(n):
    p = 2
    while True:
        if n%p!=0: return p
        else: p=nextprime(p)
def a(n): return primepi(a053669(n)) - 1 # Indranil Ghosh, May 12 2017

Scheme

(define (A276084 n) (let loop ((n n) (i 1)) (let* ((p (A000040 i)) (d (modulo n p))) (if (not (zero? d)) (- i 1) (loop (/ (- n d) p) (+ 1 i))))))

Formula

a(n) = A257993(n)-1.

Other identities. For all n >= 1:

A053589(n) = A002110(a(n)).

a(n) = A001221(A053589(n)) = A001221(A340346(n)). - Peter Munn, Jan 14 2021